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3 Forces on a Ring

  1. Jan 28, 2012 #1
    Three forces are applied to a ring (as shown in the photo) that lies on a frictionless surface in the xy plane. The ring has a mass of 100 kg. Fa=200N, Fc=240N and the angle between Fa and Fb is 135°.

    What is Fb if:

    The system is stationary?

    The system accelerates at .5 m/s2?

    For some reason I just can't get a grasp on this problem. I understand that to be stationary all forces must cancel out, but I can't figure out how with Fc having a stronger pull in the x direction than Fa how Fb (which appears to go straight down) can stop Fc.
     

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    Last edited: Jan 28, 2012
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  3. Jan 28, 2012 #2

    tiny-tim

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    hi zaper! :wink:
    you aren't told the angle between Fc and the other two …

    so you can make it anything you like! :smile:
     
  4. Jan 28, 2012 #3
    So Basically to not move I should make the angle between Fb and Fc 135 as well which means that Fa and Fc will cancel x-wise and Fb will have to be 2*Fa*cos(45)?
     
  5. Jan 28, 2012 #4

    tiny-tim

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    no, because Fa is 200 N and Fc is 240 N :redface:
     
  6. Jan 28, 2012 #5
    Oh yeah. So then 200cos(45)=240cos(x) so Fc is at 53.9. This means Fb=200sin(45)+240sin(53.9)which is 335.3
     
    Last edited: Jan 28, 2012
  7. Jan 28, 2012 #6

    tiny-tim

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    no, not 45°, you need to use an unknown θ :wink:
     
  8. Jan 28, 2012 #7
    Yeah I edited my previous post so hopefully it's correct. I'm sorry. My brain is not working this morning
     
  9. Jan 28, 2012 #8

    tiny-tim

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    ahh! :smile:
    yes! (i haven't checked the figures, but …) that method looks fine :wink:
     
  10. Jan 28, 2012 #9
    Ok so that solves the first part. Now for the second part since it goes .5m/s2 in the x direction and the ring is 100 kg then Fc is 50 N greater than Fa in the x direction so 200cos(45)=240cos(x)-50?
     
  11. Jan 28, 2012 #10
    If this method is correct then I get that Fc is at 37.1. This means that Fb=200sin(45)+240sin(37.1) which is 286.2
     
  12. Jan 28, 2012 #11

    tiny-tim

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    yes, but you'll need a y equation also :wink:
     
  13. Jan 28, 2012 #12
    I have that in my last post I believe
     
  14. Jan 28, 2012 #13

    tiny-tim

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    ahh! :redface:
    yes, that looks fine too :smile:
     
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